假设我有一个数据表,其中包含一个人的收入、工作时间和id的信息
我想计算每小时收入iph
,然后计算每个人随时间的收入发展(iphd
(。
在最后的数据表中,我想存储两个变量iph
和iphd
。
data <- data.table(
income = c(100, 120, 140, 205, 200, 220),
hours = c( 10, 11, 12, 18, 17, 21),
id = c( 1, 1, 1, 2, 2, 2)
)
(data
[, iph := income / hours]
[, iphd := c(NA, diff(iph)), by = id])[]
由于习惯了基于R的within
函数,我想在同一表达式中的定义之后立即访问iph。类似于:
# Trial no. 1
data[,
`:=`(
iph := income / hours,
iphd := c(NA, diff(iph))),
by = id][]
# Trial no. 2
data[, `:=`({
iph = income / hours
iphd = c(NA, diff(iph))
}), by = id][]
# Trial no. 3
data[, .({
iph = income / hours
iphd = c(NA, diff(iph))
}), by = id][]
然而,这些解决方案都不起作用
除了我上面建议的两步方法之外,还有其他方法可以做到这一点吗?
计算{...}
和列表中的返回结果
data[, c("iph", "iphd") := {
iph <- income / hours
iphd <- c(NA, diff(iph))
list(iph,iphd)
}, by = id]
# income hours id iph iphd
# 1: 100 10 1 10.00000 NA
# 2: 120 11 1 10.90909 0.9090909
# 3: 140 12 1 11.66667 0.7575758
# 4: 205 18 2 11.38889 NA
# 5: 200 17 2 11.76471 0.3758170
# 6: 220 21 2 10.47619 -1.2885154
不带大括号:
data[, c("iph", "iphd") := list(income / hours,
c(NA, diff(income / hours))), by = id][]